Question

which is subtracted from each 21,38,55 and 106 that the new numbers are in the proportional​

Answer 1

Answer:

What number must be subtracted from 21, 38, 55, and 106 each so that the remainders( technically differences) are proportional?

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Using the J programming language:

Brute Force:

What integers divided into 21, 38, 55 & 106 all have the same remainders?

>:I.1=(#@~.)"1(>:i.21)|/21 38 55 106

1 17

What is the remainder when dividing all four integers by 17.

17|21 38 55 106

4 4 4 4

So we need to subtract 4 from all four integers to make them all proportional by integers.

What are the ratios after subtracting 4 and dividing by 17?

17%~21 38 55 106-4

1 2 3 6

So the proportions are 1, 2, 3, & 6.

Answer 2

4 must be subtracted from each 21 , 38 , 55 and 106 that the new numbers are in the proportional

Given :

The numbers 21 , 38 , 55 and 106

To find :

The must be subtracted from each 21 , 38 , 55 and 106 that the new numbers are in the proportional

Solution :

Step 1 of 2 :

Form the equation to find the number

Let x must be subtracted from each 21 , 38 , 55 and 106 that the new numbers are in the proportional

By the given condition

$\displaystyle \sf (21 - x):(38 - x) = (55 - x):(106 - x)$

Step 2 of 2 :

Find the number

$\displaystyle \sf (21 - x):(38 - x) = (55 - x):(106 - x)$

$\displaystyle \sf \implies \frac{ (21 - x)}{(38 - x) }= \frac{(55 - x)}{(106 - x)}$

$\displaystyle \sf{ \implies }(21 - x)(106 - x) = (38 - x)(55 - x)$

$\displaystyle \sf{ \implies }2226 - 106x - 21x + {x}^{2} = 2090 - 38x - 55x + {x}^{2}$

$\displaystyle \sf{ \implies }2226 - 127x = 2090 - 93x$

$\displaystyle \sf{ \implies } - 127x + 93x = 2090 - 2226$

$\displaystyle \sf{ \implies } - 34x = - 136$

$\displaystyle \sf{ \implies }x = \frac{ - 136}{ - 34}$

$\displaystyle \sf{ \implies }x = 4$

Hence 4 must be subtracted from each 21 , 38 , 55 and 106 that the new numbers are in the proportional

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